Question

A balanced three-phase load is connected to a $\mathbf{4}\mathbf{.}\mathbf{16}\mathit{k}\mathit{V}$, three-phase, four Wire, grounded-wye dedicated distribution feeder. The load can be modeled by an impedance of${\mathit{Z}}_{\mathbf{L}}\mathbf{=}\left(4.7+j9\right)\mathit{\Omega }\mathbf{/}\mathit{p}\mathit{h}\mathit{a}\mathit{s}\mathit{e}$wye-connected. The impedance of the phase conductors is $\left(0.3+j1\right)\mathit{\Omega }$. Determine the following by using the phase$\mathit{A}$ to neutral voltage as a reference and assume positive phase sequence:

(a) Line currents for phases and.

(b) Line-to-neutral voltages for all three phases $\mathit{A}\mathbf{,}\mathit{B}$at the load$\mathit{C}$.

(c) Apparent, active, and reactive power dissipated per phase, and for all

three phases in the load.

(d) Active power losses per phase and for all three phases in the phase

conductors.

Short answer

(a)The line current for phases$A,B$,and$C$ are$214.82\angle -63.43°,$$214.82\angle -183.43°\text{\hspace{0.17em}A}$ , and $214.82\angle -303.43°\text{\hspace{0.17em}A}$,respectively.

(b) The line to neutral voltage for phases$A,B$,and$C$ is$2179.5\angle -1.01\text{\hspace{0.17em}V}$,$2179.5\angle -121.01\text{\hspace{0.17em}V}$ , $2179.5\angle -241.01\text{\hspace{0.17em}V}$, respectively.

(c)The apparent, active,and reactive power forthesingle phases are$468.2\angle 62.42°,$

$216.77\text{\hspace{0.17em}kW}$,and $414.99\text{\hspace{0.17em}kVAR}$,respectively.

The apparent, active,and reactive power forthethree phasesare$1404.6\text{\hspace{0.17em}kVA}$,$650.31\text{\hspace{0.17em}kW}$ and$1244.97\text{\hspace{0.17em}kVAR}$ ,respectively.

(d) The active power loss forasingle-phase is and for three phases is$41.53\text{\hspace{0.17em}kW}$.

Step-by-step solution

Write all the values given in the question.

Line to line voltage,$V_{LL}=4.16\text{\hspace{0.17em}kV}$.

Load impedance,$Z_L=\left(4.7+j9\right)\Omega /\text{phase}$.

The phase conductor impedance,$Z_{load}=\left(0.3+j1\right)\Omega$.

Determine the formulas to calculate the line current, line to neutral voltage, apparent, active, and reactive power for the three phases.

The expression fortheline to neutral voltage iscalculated as follows:

${\mathit{V}}_{\mathbf{AN}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{LL}}}{\sqrt{\mathbf{3}}}\mathbf{\angle }\mathbf{0}\mathbf{°}$ …… (1)

The total impedance iscalculated as follows:

$\mathit{Z}\mathbf{=}{\mathit{Z}}_{\mathbf{L}}\mathbf{+}{\mathit{Z}}_{\mathbf{load}}$ …… (2)

The line current for phase A iscalculated as follows:

${\mathit{I}}_{\mathbf{A}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{AN}}}{\mathbf{Z}}$ …… (3)

Forabalancedthree-phase system, all the line currents areseparated by the phase shift of$\mathbf{120}\mathbf{°}$.

The line to neutral voltage for the phase A iscalculated as follows:

localid="1656747793742" ${\mathbf{\left(}{\mathbf{V}}_{\mathbf{AN}}\mathbf{\right)}}_{\mathbf{Load}}\mathbf{=}{\mathit{V}}_{\mathbf{AN}}\mathbf{-}{\mathit{I}}_{\mathbf{A}}{\mathit{Z}}_{\mathbf{Load}}$ …… (4)

Forabalancedthree-phase system,linesto neutral voltageareseparated by the phase shift of$\mathbf{120}\mathbf{°}$.

The expression for the apparent power iscalculated as follows:

localid="1656747829505" ${\mathit{S}}_{\mathbf{1}\mathbf{\varphi }}\mathbf{=}{\mathbf{\left(}{\mathbf{V}}_{\mathbf{AN}}\mathbf{\right)}}_{\mathbf{Load}}{\mathit{I}}_{\mathbf{A}}^{\mathbf{*}}$ …… (5)

The apparent power for three-phasesis calculated as follows:

localid="1656747847685" ${\mathit{S}}_{\mathbf{3}\mathbf{\varphi }}\mathbf{=}\mathbf{3}{\mathit{S}}_{\mathbf{1}\mathbf{\varphi }}$ …… (6)

The expression for the active power iscalculated as follows:

localid="1656747885115" ${\mathit{P}}_{\mathbf{1}\mathbf{\varphi }}\mathbf{=}{\mathit{S}}_{\mathbf{1}\mathbf{\varphi }}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}$ …… (7)

Here,$\mathit{\theta }$is the angle between the line voltage and current.

The expression for the reactive power iscalculated as follows:

localid="1656747902723" ${\mathit{Q}}_{\mathbf{1}\mathbf{\varphi }}\mathbf{=}{\mathit{S}}_{\mathbf{1}\mathbf{\varphi }}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}$ …… (8)

The active power for all three-phase iscalculated as follows:

${\mathit{P}}_{\mathbf{3}\mathbf{\varphi }}\mathbf{=}\mathbf{3}{\mathit{P}}_{\mathbf{1}\mathbf{\varphi }}$ …… (9)

The reactive power for all three-phase iscalculated as follows:

${\mathit{Q}}_{\mathbf{3}\mathbf{\varphi }}\mathbf{=}\mathbf{3}{\mathit{Q}}_{\mathbf{1}\mathbf{\varphi }}$ …… (10)

The active power loss for single-phase iscalculated as follows:

localid="1656747926293" ${\mathbf{\left(}{\mathbf{P}}_{\mathbf{1}\mathbf{\varphi }}\mathbf{\right)}}_{\mathbf{Load}}\mathbf{=}{\mathit{I}}^{\mathbf{2}}\mathit{R}$ …… (11)

Here,$\mathit{I}$is the line current,and$\mathit{R}$is the per phase resistance of the conductor.

The active power loss forthethree-phase iscalculated as follows:

localid="1656747944381" ${\mathit{P}}_{\mathbf{3}\mathbf{\varphi }}\mathbf{=}\mathbf{3}{\mathit{P}}_{\mathbf{1}\mathbf{\varphi }}$ …… (12)

Determine the line current for the phase A, B, andC.

(a)

Calculate the voltage between phase $A$and neutral $N$using equation (1).

$\begin{array}{l}{V}_{AN}=\frac{4.16\times 1000}{\sqrt{3}}\angle 0°\\ V_{AN}=2401.77\angle 0°\text{\hspace{0.17em}V}\end{array}$

The total impedance is the sum of the load impedance and impedance of the phase conductor (using equation 2).

$\begin{array}{l}Z=4.7+j9+0.3+j1\\ Z=5+j10\end{array}$

The line current for phase$A$can be calculated using equation (3).

$\begin{array}{l}{I}_A=\frac{2401.7\angle 0°}{5+j10}\\ I_A=214.82\angle -63.43°\end{array}$

For the balancedthree-phase system,the line current$I_B$and$I_C$ would be shifted by$120°$.

The line current $I_B$is as follows:

$\begin{array}{l}{I}_B=214.82\angle \left(-63.43-120\right)°\\ I_B=214.82\angle -183.43°\text{\hspace{0.17em}A}\end{array}$

The line current $I_C$isas follows:

$\begin{array}{l}{I}_C=214.82\angle \left(-63.43-240\right)°\\ I_C=214.82\angle -303.43°\text{\hspace{0.17em}A}\end{array}$

Hence, the line current for the phase$A,B$ , and$C$ are $214.82\angle -63.43°,$$214.82\angle -183.43°\text{\hspace{0.17em}A}$, and $214.82\angle -303.43°\text{\hspace{0.17em}A}$, respectively.

The line to neutral voltage for the phaseA,B , andC.

(b)

The line to neutral voltage for phase $A$can be calculated by equation (4).

$\begin{array}{l}{\left(V_{AN}\right)}_{Load}=2401.77\angle 0°-\left(96-j192\right)\left(0.3\_j1\right)\\ {\left(V_{AN}\right)}_{Load}=2401.77\angle -\left(220.8+j38.4\right)\\ {\left(V_{AN}\right)}_{Load}=2179.2-j38.4\\ {\left(V_{AN}\right)}_{Load}=2179.5\angle -1.01\text{\hspace{0.17em}V}\end{array}$

The line to neutral current for phase $B$isas follows:

$\begin{array}{l}{\left(V_{BN}\right)}_{Load}=2179.5\angle \left(-1.01-120\right)\text{\hspace{0.17em}V}\\ {\left(V_{BN}\right)}_{Load}=2179.5\angle -121.01\text{\hspace{0.17em}V}\end{array}$

The line to neutral current for phase$C$ isas follows:

$\begin{array}{l}{\left(V_{CN}\right)}_{Load}=2179.5\angle \left(-1.01-240\right)\text{\hspace{0.17em}V}\\ {\left(V_{CN}\right)}_{Load}=2179.5\angle -241.01\text{\hspace{0.17em}V}\end{array}$

Hence the line to neutral voltage for phase ,$A,B$and$C$ are$2179.5\angle -1.01\text{\hspace{0.17em}V}$, $2179.5\angle -121.01\text{\hspace{0.17em}V}$, $2179.5\angle -241.01\text{\hspace{0.17em}V}$,respectively.

Calculate the apparent, active and reactive power dissipated in all three phases. 

(c)

Calculate apparent power for single-phase using equation (5).

$\begin{array}{l}{S}_{1\varphi }=\left(2179.5\angle -1.01°\right)\left(214.82\angle 63.43°\right)\\ S_{1\varphi }=468.2\angle 62.42°\end{array}$

Calculate the active power for single-phase using equation (7).

$\begin{array}{l}{P}_{1\varphi }=486.2\cos \left(62.42\right)\\ P_{1\varphi }=216.77\text{\hspace{0.17em}kW}\end{array}$

Calculate the reactive power for single-phase using equation (8).

$\begin{array}{l}{Q}_{1\varphi }=468.2\times \sin \left(62.42\right)\\ Q_{1\varphi }=414.99\text{\hspace{0.17em}kVAR}\end{array}$

Calculate the apparent power for three phases using equation (6).

$\begin{array}{l}{S}_{3\varphi }=3\times 468.2\angle 62.42°\\ S_{3\varphi }=1404.6\text{\hspace{0.17em}kVA}\end{array}$

Calculate the active power for threephases using equation (9).

$\begin{array}{l}{P}_{3\varphi }=3\times 216.77\\ P_{3\varphi }=650.31\text{\hspace{0.17em}kW}\end{array}$

Calculate the reactive power for three phases using equation (10).

$\begin{array}{l}{Q}_{1\varphi }=3\times 414.99\\ Q_{1\varphi }=1244.97\text{\hspace{0.17em}kVAR}\end{array}$

Hence the apparent, active,and reactive power for the single-phase is

$468.2\angle 62.42°,$$216.77\text{\hspace{0.17em}kW}$,and$414.99\text{\hspace{0.17em}kVAR}$ ,respectively.

The apparent, active, and reactive power for three phases is $1404.6\text{\hspace{0.17em}kVA}$,$650.31\text{\hspace{0.17em}kW}$, and $1244.97\text{\hspace{0.17em}kVAR}$, respectively.

Calculate the active for loss for all per phase and three phases.

(d)

Calculate the active power loss for the per phase using equation (11),

$\begin{array}{l}{\left(P_{1\varphi }\right)}_{Load}={214.82}^2\times 0.3\\ {\left(P_{1\varphi }\right)}_{Load}=13.84\text{\hspace{0.17em}kW}\end{array}$

Calculate the active power loss for the three phases using equation (12),

$\begin{array}{l}{P}_{3\varphi }=3\times 13.84\\ P_{3\varphi }=41.53\text{\hspace{0.17em}kW}\end{array}$

Hence,the active power loss for single-phase is $13.84\text{\hspace{0.17em}kW}$and for three phases is$41.53\text{\hspace{0.17em}kW}$ .